Mixing
Using dominated convergence theorem to move limit inside the integral
Clash Royale CLAN TAG#URR8PPP
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I have to prove that the following limit goes to $0$:
$$lim_ntoinfty int_-infty^infty left|u(omega)bigg(frac2n^2omega^2(1-cos(omega/n))-1bigg)right|^2dw$$
where $u in L^2(mathbbR)$.
I want to move the limit inside to that I can substitute:
$$cos(1/t) = 1 - frac12t^2 + frac14!t^4 - text... $$
And the proof is completed.
I was thinking about using the dominated convergence theorem, but I am not sure how to use it. As a dominating sequence I would use that:
$$1-cos(t) leq t^2/2 $$
But then, what else should I make sure of?
integration limits convergence
edited Jul 30 at 20:53
ComplexYetTrivial
2,637624
asked Jul 30 at 18:21
giovanni_13
566
add a comment |Â
up vote
2
down vote
favorite
I have to prove that the following limit goes to $0$:
$$lim_ntoinfty int_-infty^infty left|u(omega)bigg(frac2n^2omega^2(1-cos(omega/n))-1bigg)right|^2dw$$
where $u in L^2(mathbbR)$.
I want to move the limit inside to that I can substitute:
$$cos(1/t) = 1 - frac12t^2 + frac14!t^4 - text... $$
And the proof is completed.
I was thinking about using the dominated convergence theorem, but I am not sure how to use it. As a dominating sequence I would use that:
$$1-cos(t) leq t^2/2 $$
But then, what else should I make sure of?
integration limits convergence
edited Jul 30 at 20:53
ComplexYetTrivial
2,637624
asked Jul 30 at 18:21
giovanni_13
566
$dt$? I don't see $t$ there :c
– Gonzalo Benavides
Jul 30 at 18:27
I edited the post to add the missing information
– giovanni_13
Jul 30 at 19:11
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I have to prove that the following limit goes to $0$:
$$lim_ntoinfty int_-infty^infty left|u(omega)bigg(frac2n^2omega^2(1-cos(omega/n))-1bigg)right|^2dw$$
where $u in L^2(mathbbR)$.
I want to move the limit inside to that I can substitute:
$$cos(1/t) = 1 - frac12t^2 + frac14!t^4 - text... $$
And the proof is completed.
I was thinking about using the dominated convergence theorem, but I am not sure how to use it. As a dominating sequence I would use that:
$$1-cos(t) leq t^2/2 $$
But then, what else should I make sure of?
integration limits convergence
edited Jul 30 at 20:53
ComplexYetTrivial
2,637624
asked Jul 30 at 18:21
giovanni_13
566
I have to prove that the following limit goes to $0$:
$$lim_ntoinfty int_-infty^infty left|u(omega)bigg(frac2n^2omega^2(1-cos(omega/n))-1bigg)right|^2dw$$
where $u in L^2(mathbbR)$.
I want to move the limit inside to that I can substitute:
$$cos(1/t) = 1 - frac12t^2 + frac14!t^4 - text... $$
And the proof is completed.
I was thinking about using the dominated convergence theorem, but I am not sure how to use it. As a dominating sequence I would use that:
$$1-cos(t) leq t^2/2 $$
But then, what else should I make sure of?
integration limits convergence
edited Jul 30 at 20:53
ComplexYetTrivial
2,637624
asked Jul 30 at 18:21
giovanni_13
566
edited Jul 30 at 20:53
ComplexYetTrivial
2,637624
edited Jul 30 at 20:53
ComplexYetTrivial
2,637624
edited Jul 30 at 20:53
ComplexYetTrivial
2,637624
2,637624
asked Jul 30 at 18:21
giovanni_13
566
asked Jul 30 at 18:21
giovanni_13
566
asked Jul 30 at 18:21
giovanni_13
566
566
$dt$? I don't see $t$ there :c
– Gonzalo Benavides
Jul 30 at 18:27
I edited the post to add the missing information
– giovanni_13
Jul 30 at 19:11
add a comment |Â
$dt$? I don't see $t$ there :c
– Gonzalo Benavides
Jul 30 at 18:27
I edited the post to add the missing information
– giovanni_13
Jul 30 at 19:11
$dt$? I don't see $t$ there :c
– Gonzalo Benavides
Jul 30 at 18:27
$dt$? I don't see $t$ there :c
– Gonzalo Benavides
Jul 30 at 18:27
I edited the post to add the missing information
– giovanni_13
Jul 30 at 19:11
I edited the post to add the missing information
– giovanni_13
Jul 30 at 19:11
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
We can use your inequality to show that
$$ 0 leq frac2(1-cos(t))t^2 leq 1$$
holds for $t in mathbbR$ .
If we define $f_n colon mathbbR to mathbbR$ to be your integrand, i.e.
$$ f_n(omega) = |u(omega)|^2 left[1 - frac2 (1-cos(omega /n))(omega/n)^2right]^2 , , $$
we immediately obtain $|f_n(omega)| leq |u(omega)|^2$ for all $omega in mathbbR$ and every $n in mathbbN$ from the above inequality.
We take $|u|^2$ as the dominating function from the theorem. It is integrable, as we have $u in L^2 (mathbbR)$ . Therefore the dominated convergence theorem allows us to interchange the limit and the integration. But now we are done, since $(f_n)_nin mathbbN$ converges to zero pointwise almost everywhere (as mentioned in your question).
answered Jul 30 at 20:43
ComplexYetTrivial
2,637624
that's a very good solution, thank you! But why would it be wrong to say that: $lim_ntoinfty int_-infty^infty left|u(omega)bigg(frac2n^2omega^2(1-cos(omega/n))-1bigg)right|^2dw leq lim_ntoinfty int_-infty^infty left|u(omega)bigg(bigg(frac2n^2omega^2fracomega^22n^2bigg)-1bigg)right|^2dw$ $= lim_ntoinfty int_-infty^infty left|u(omega)(1-1)right|^2dw = 0$
– giovanni_13
Jul 30 at 21:22
1
@giovanni_13 Because of the absolute value. We have $|2(1-cos(x))x^-2 -1| = 1 - 2(1-cos(x))x^-2$ and now you can use the inequality. But it only shows that this quantity is $geq 0$ instead of $leq 0$, so you cannot conclude that all integrals vanish. One must be very careful when applying inequalities to expressions with absolute value bars.
– ComplexYetTrivial
Jul 30 at 21:42
Thank you very much, now everything is really clear.
– giovanni_13
Jul 30 at 21:47
just a doubt: should I also make sure that $f_n(x)$ is a sequence of measurable functions $forall n in mathbbN$
– giovanni_13
Jul 31 at 7:32
@giovanni_13 That is correct. Since $u in L^2 (mathbbR)$ is measurable, and $omega mapsto 1-2(2-cos(omega/n)) (omega/n)^-2$ is even continuous, their product and its absolute value squared are indeed measurable.
– ComplexYetTrivial
Jul 31 at 13:09
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
We can use your inequality to show that
$$ 0 leq frac2(1-cos(t))t^2 leq 1$$
holds for $t in mathbbR$ .
If we define $f_n colon mathbbR to mathbbR$ to be your integrand, i.e.
$$ f_n(omega) = |u(omega)|^2 left[1 - frac2 (1-cos(omega /n))(omega/n)^2right]^2 , , $$
we immediately obtain $|f_n(omega)| leq |u(omega)|^2$ for all $omega in mathbbR$ and every $n in mathbbN$ from the above inequality.
We take $|u|^2$ as the dominating function from the theorem. It is integrable, as we have $u in L^2 (mathbbR)$ . Therefore the dominated convergence theorem allows us to interchange the limit and the integration. But now we are done, since $(f_n)_nin mathbbN$ converges to zero pointwise almost everywhere (as mentioned in your question).
answered Jul 30 at 20:43
ComplexYetTrivial
2,637624
that's a very good solution, thank you! But why would it be wrong to say that: $lim_ntoinfty int_-infty^infty left|u(omega)bigg(frac2n^2omega^2(1-cos(omega/n))-1bigg)right|^2dw leq lim_ntoinfty int_-infty^infty left|u(omega)bigg(bigg(frac2n^2omega^2fracomega^22n^2bigg)-1bigg)right|^2dw$ $= lim_ntoinfty int_-infty^infty left|u(omega)(1-1)right|^2dw = 0$
– giovanni_13
Jul 30 at 21:22
1
@giovanni_13 Because of the absolute value. We have $|2(1-cos(x))x^-2 -1| = 1 - 2(1-cos(x))x^-2$ and now you can use the inequality. But it only shows that this quantity is $geq 0$ instead of $leq 0$, so you cannot conclude that all integrals vanish. One must be very careful when applying inequalities to expressions with absolute value bars.
– ComplexYetTrivial
Jul 30 at 21:42
Thank you very much, now everything is really clear.
– giovanni_13
Jul 30 at 21:47
just a doubt: should I also make sure that $f_n(x)$ is a sequence of measurable functions $forall n in mathbbN$
– giovanni_13
Jul 31 at 7:32
@giovanni_13 That is correct. Since $u in L^2 (mathbbR)$ is measurable, and $omega mapsto 1-2(2-cos(omega/n)) (omega/n)^-2$ is even continuous, their product and its absolute value squared are indeed measurable.
– ComplexYetTrivial
Jul 31 at 13:09
add a comment |Â
up vote
1
down vote
accepted
We can use your inequality to show that
$$ 0 leq frac2(1-cos(t))t^2 leq 1$$
holds for $t in mathbbR$ .
If we define $f_n colon mathbbR to mathbbR$ to be your integrand, i.e.
$$ f_n(omega) = |u(omega)|^2 left[1 - frac2 (1-cos(omega /n))(omega/n)^2right]^2 , , $$
we immediately obtain $|f_n(omega)| leq |u(omega)|^2$ for all $omega in mathbbR$ and every $n in mathbbN$ from the above inequality.
We take $|u|^2$ as the dominating function from the theorem. It is integrable, as we have $u in L^2 (mathbbR)$ . Therefore the dominated convergence theorem allows us to interchange the limit and the integration. But now we are done, since $(f_n)_nin mathbbN$ converges to zero pointwise almost everywhere (as mentioned in your question).
answered Jul 30 at 20:43
ComplexYetTrivial
2,637624
that's a very good solution, thank you! But why would it be wrong to say that: $lim_ntoinfty int_-infty^infty left|u(omega)bigg(frac2n^2omega^2(1-cos(omega/n))-1bigg)right|^2dw leq lim_ntoinfty int_-infty^infty left|u(omega)bigg(bigg(frac2n^2omega^2fracomega^22n^2bigg)-1bigg)right|^2dw$ $= lim_ntoinfty int_-infty^infty left|u(omega)(1-1)right|^2dw = 0$
– giovanni_13
Jul 30 at 21:22
1
@giovanni_13 Because of the absolute value. We have $|2(1-cos(x))x^-2 -1| = 1 - 2(1-cos(x))x^-2$ and now you can use the inequality. But it only shows that this quantity is $geq 0$ instead of $leq 0$, so you cannot conclude that all integrals vanish. One must be very careful when applying inequalities to expressions with absolute value bars.
– ComplexYetTrivial
Jul 30 at 21:42
Thank you very much, now everything is really clear.
– giovanni_13
Jul 30 at 21:47
just a doubt: should I also make sure that $f_n(x)$ is a sequence of measurable functions $forall n in mathbbN$
– giovanni_13
Jul 31 at 7:32
@giovanni_13 That is correct. Since $u in L^2 (mathbbR)$ is measurable, and $omega mapsto 1-2(2-cos(omega/n)) (omega/n)^-2$ is even continuous, their product and its absolute value squared are indeed measurable.
– ComplexYetTrivial
Jul 31 at 13:09
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
We can use your inequality to show that
$$ 0 leq frac2(1-cos(t))t^2 leq 1$$
holds for $t in mathbbR$ .
If we define $f_n colon mathbbR to mathbbR$ to be your integrand, i.e.
$$ f_n(omega) = |u(omega)|^2 left[1 - frac2 (1-cos(omega /n))(omega/n)^2right]^2 , , $$
we immediately obtain $|f_n(omega)| leq |u(omega)|^2$ for all $omega in mathbbR$ and every $n in mathbbN$ from the above inequality.
We take $|u|^2$ as the dominating function from the theorem. It is integrable, as we have $u in L^2 (mathbbR)$ . Therefore the dominated convergence theorem allows us to interchange the limit and the integration. But now we are done, since $(f_n)_nin mathbbN$ converges to zero pointwise almost everywhere (as mentioned in your question).
answered Jul 30 at 20:43
ComplexYetTrivial
2,637624
We can use your inequality to show that
$$ 0 leq frac2(1-cos(t))t^2 leq 1$$
holds for $t in mathbbR$ .
If we define $f_n colon mathbbR to mathbbR$ to be your integrand, i.e.
$$ f_n(omega) = |u(omega)|^2 left[1 - frac2 (1-cos(omega /n))(omega/n)^2right]^2 , , $$
we immediately obtain $|f_n(omega)| leq |u(omega)|^2$ for all $omega in mathbbR$ and every $n in mathbbN$ from the above inequality.
We take $|u|^2$ as the dominating function from the theorem. It is integrable, as we have $u in L^2 (mathbbR)$ . Therefore the dominated convergence theorem allows us to interchange the limit and the integration. But now we are done, since $(f_n)_nin mathbbN$ converges to zero pointwise almost everywhere (as mentioned in your question).
answered Jul 30 at 20:43
ComplexYetTrivial
2,637624
answered Jul 30 at 20:43
ComplexYetTrivial
2,637624
answered Jul 30 at 20:43
ComplexYetTrivial
2,637624
answered Jul 30 at 20:43
ComplexYetTrivial
2,637624
2,637624
that's a very good solution, thank you! But why would it be wrong to say that: $lim_ntoinfty int_-infty^infty left|u(omega)bigg(frac2n^2omega^2(1-cos(omega/n))-1bigg)right|^2dw leq lim_ntoinfty int_-infty^infty left|u(omega)bigg(bigg(frac2n^2omega^2fracomega^22n^2bigg)-1bigg)right|^2dw$ $= lim_ntoinfty int_-infty^infty left|u(omega)(1-1)right|^2dw = 0$
– giovanni_13
Jul 30 at 21:22
1
@giovanni_13 Because of the absolute value. We have $|2(1-cos(x))x^-2 -1| = 1 - 2(1-cos(x))x^-2$ and now you can use the inequality. But it only shows that this quantity is $geq 0$ instead of $leq 0$, so you cannot conclude that all integrals vanish. One must be very careful when applying inequalities to expressions with absolute value bars.
– ComplexYetTrivial
Jul 30 at 21:42
Thank you very much, now everything is really clear.
– giovanni_13
Jul 30 at 21:47
just a doubt: should I also make sure that $f_n(x)$ is a sequence of measurable functions $forall n in mathbbN$
– giovanni_13
Jul 31 at 7:32
@giovanni_13 That is correct. Since $u in L^2 (mathbbR)$ is measurable, and $omega mapsto 1-2(2-cos(omega/n)) (omega/n)^-2$ is even continuous, their product and its absolute value squared are indeed measurable.
– ComplexYetTrivial
Jul 31 at 13:09
add a comment |Â
that's a very good solution, thank you! But why would it be wrong to say that: $lim_ntoinfty int_-infty^infty left|u(omega)bigg(frac2n^2omega^2(1-cos(omega/n))-1bigg)right|^2dw leq lim_ntoinfty int_-infty^infty left|u(omega)bigg(bigg(frac2n^2omega^2fracomega^22n^2bigg)-1bigg)right|^2dw$ $= lim_ntoinfty int_-infty^infty left|u(omega)(1-1)right|^2dw = 0$
– giovanni_13
Jul 30 at 21:22
1
@giovanni_13 Because of the absolute value. We have $|2(1-cos(x))x^-2 -1| = 1 - 2(1-cos(x))x^-2$ and now you can use the inequality. But it only shows that this quantity is $geq 0$ instead of $leq 0$, so you cannot conclude that all integrals vanish. One must be very careful when applying inequalities to expressions with absolute value bars.
– ComplexYetTrivial
Jul 30 at 21:42
Thank you very much, now everything is really clear.
– giovanni_13
Jul 30 at 21:47
just a doubt: should I also make sure that $f_n(x)$ is a sequence of measurable functions $forall n in mathbbN$
– giovanni_13
Jul 31 at 7:32
@giovanni_13 That is correct. Since $u in L^2 (mathbbR)$ is measurable, and $omega mapsto 1-2(2-cos(omega/n)) (omega/n)^-2$ is even continuous, their product and its absolute value squared are indeed measurable.
– ComplexYetTrivial
Jul 31 at 13:09
that's a very good solution, thank you! But why would it be wrong to say that: $lim_ntoinfty int_-infty^infty left|u(omega)bigg(frac2n^2omega^2(1-cos(omega/n))-1bigg)right|^2dw leq lim_ntoinfty int_-infty^infty left|u(omega)bigg(bigg(frac2n^2omega^2fracomega^22n^2bigg)-1bigg)right|^2dw$ $= lim_ntoinfty int_-infty^infty left|u(omega)(1-1)right|^2dw = 0$
– giovanni_13
Jul 30 at 21:22
that's a very good solution, thank you! But why would it be wrong to say that: $lim_ntoinfty int_-infty^infty left|u(omega)bigg(frac2n^2omega^2(1-cos(omega/n))-1bigg)right|^2dw leq lim_ntoinfty int_-infty^infty left|u(omega)bigg(bigg(frac2n^2omega^2fracomega^22n^2bigg)-1bigg)right|^2dw$ $= lim_ntoinfty int_-infty^infty left|u(omega)(1-1)right|^2dw = 0$
– giovanni_13
Jul 30 at 21:22
1
1
@giovanni_13 Because of the absolute value. We have $|2(1-cos(x))x^-2 -1| = 1 - 2(1-cos(x))x^-2$ and now you can use the inequality. But it only shows that this quantity is $geq 0$ instead of $leq 0$, so you cannot conclude that all integrals vanish. One must be very careful when applying inequalities to expressions with absolute value bars.
– ComplexYetTrivial
Jul 30 at 21:42
@giovanni_13 Because of the absolute value. We have $|2(1-cos(x))x^-2 -1| = 1 - 2(1-cos(x))x^-2$ and now you can use the inequality. But it only shows that this quantity is $geq 0$ instead of $leq 0$, so you cannot conclude that all integrals vanish. One must be very careful when applying inequalities to expressions with absolute value bars.
– ComplexYetTrivial
Jul 30 at 21:42
Thank you very much, now everything is really clear.
– giovanni_13
Jul 30 at 21:47
Thank you very much, now everything is really clear.
– giovanni_13
Jul 30 at 21:47
just a doubt: should I also make sure that $f_n(x)$ is a sequence of measurable functions $forall n in mathbbN$
– giovanni_13
Jul 31 at 7:32
just a doubt: should I also make sure that $f_n(x)$ is a sequence of measurable functions $forall n in mathbbN$
– giovanni_13
Jul 31 at 7:32
@giovanni_13 That is correct. Since $u in L^2 (mathbbR)$ is measurable, and $omega mapsto 1-2(2-cos(omega/n)) (omega/n)^-2$ is even continuous, their product and its absolute value squared are indeed measurable.
– ComplexYetTrivial
Jul 31 at 13:09
@giovanni_13 That is correct. Since $u in L^2 (mathbbR)$ is measurable, and $omega mapsto 1-2(2-cos(omega/n)) (omega/n)^-2$ is even continuous, their product and its absolute value squared are indeed measurable.
– ComplexYetTrivial
Jul 31 at 13:09
add a comment |Â
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$dt$? I don't see $t$ there :c
– Gonzalo Benavides
Jul 30 at 18:27
I edited the post to add the missing information
– giovanni_13
Jul 30 at 19:11